Does Cauchy Test Fail If $\lim_{x\to\infty}\int_x^{2x}f(t)dt = 0$?

[no_alone]

Homework Statement

True Or False
if f(x) continuous in [tex][a,\infty][/tex] and [tex]\lim_{x\to\infty}\int_x^{2x}f(t)dt = 0 [/tex]
Then [tex]\int_a^\infty f(x)dx [/tex] converge

Homework Equations

Anything from calc 1 and 2

The Attempt at a Solution

Actually I'm really stuck..
My main motive is to try and show that if [tex]\lim_{x\to\infty}\int_x^{2x}f(t)dt = 0 [/tex] there is no way that Cauchy test can fail,
I assume Cauchy test fail and that [tex]\lim_{x\to\infty}\int_x^{2x}f(t)dt = 0 [/tex] and try to show that there is no option for this to happen.
First I tried on f(x) that can be positive and negative .. didn't made it, Then I tried on f(x) that is only positive also didn't made it..
Maybe its false, But I didn't found any function that does not converge and also [tex]\lim_{x\to\infty}\int_x^{2x}f(t)dt = 0 [/tex]
Thank you.

 

[Dick]

Well, let's try and make such a function. Suppose we make the integral over [a,2a]~1/2, the integral over [2a,4a]~1/3, the integral over [4a,8a]~1/4. Then integral over all t should diverge like a harmonic series, right? Now, what must the values of such a function look like? Over [a,2a] it should be about 1/(2a). Over [2a,4a] it should be about 1/(3*(2a)). Over [4a,8a] like about 1/(4*(4a)). Etc, etc. That sort of looks like 1/(log(t)*t) to me.

 

[no_alone]

Thank you Dick
I didn't thought about series all I had in mind were function..
I tried and tried , even when I tried to build a function like you did by [a,2a],[4a,8a], I did it by building it by other function, I never imagine building a function with a series
Thank you.